Atcoder Regular Contest 145

By xiaruize

A - AB Palindrome

Stop Learning useless algorithms. --Um_nik

当你在想dp，想贪心，想hash时，都不会的Beginner已经AC\color{green}{AC}了！！！

先特判n=2n=2的情况

其他的s[0]=As[0]=A且s[n−1]=Bs[n-1]=B的是NoNo

其他都是YesYes

Code
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// Problem: A - AB Palindrome
// Contest: AtCoder - AtCoder Regular Contest 145
// URL: https://atcoder.jp/contests/arc145/tasks/arc145_a
// Memory Limit: 1024 MB
// Time Limit: 2000 ms
//
// Powered by CP Editor (https://cpeditor.org)

/*
    Name:
    Author: xiaruize
    Date:
*/
#include <bits/stdc++.h>
using namespace std;
#define int long long
#define ull unsigned long long
#define ALL(a) (a).begin(), (a).end()
#define pb push_back
#define mk make_pair
#define pii pair<int, int>
#define pis pair<int, string>
#define sec second
#define fir first
#define sz(a) int((a).size())
#define rep(i, x, y) for (int i = x; i <= y; i++)
#define repp(i, x, y) for (int i = x; i >= y; i--)
#define Yes cout << "Yes" << endl
#define YES cout << "YES" << endl
#define No cout << "No" << endl
#define NO cout << "NO" << endl
#define debug(x) cerr << #x << ": " << x << endl
#define double long double
const int INF = 0x3f3f3f3f;
const int MOD = 1000000007;
const int N = 2e5 + 10;

// bool st;
string s;
int n;
// bool en;

signed main()
{
    // freopen(".in","r",stdin);
    // freopen(".out","w",stdout);
    // cerr<<(&en-&st)/1024.0/1024.0<<endl;
    // auto t_1=chrono::high_resolution_clock::now();
    cin >> n;
    cin >> s;
    if (n == 2)
    {
        if (s == "AA" || s == "BB")
            Yes;
        else
            No;
        return 0;
    }
    if (s[0] == 'A' && s[n - 1] == 'B')
        No;
    else
        Yes;
    // auto t_2=chrono::high_resolution_clock::now();
    // cout <<". Elapsed (ms): " << chrono::duration_cast<chrono::milliseconds>(t_2 - t_1).count() << endl;
    return 0;
}

B - AB Game

1≤N,A,B≤10181\leq N,A,B\leq 10^{18}明显数学，分类讨论

  • n<an<a，答案为00
  • b>ab>a, 答案为n−a+1n-a+1
  • 其他情况 答案为 na+(na−1)⋅(b−1)+min(nmod  a,b−1)\frac{n}{a}+(\frac{n}{a}-1)\cdot (b-1)+min(n\mod a,b-1)

前两个情况是显然的，这里解释一下第三个

首先，第一次一定取尽可能多的aa使得剩下的值小于bb

形式化表示就是求

{1≤i≤nimod  a<b\begin{cases} 1\leq i \leq n\\ i \mod a < b \end{cases}

然后就可以数学计算啦

Code
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// Problem: B - AB Game
// Contest: AtCoder - AtCoder Regular Contest 145
// URL: https://atcoder.jp/contests/arc145/tasks/arc145_b
// Memory Limit: 1024 MB
// Time Limit: 2000 ms
//
// Powered by CP Editor (https://cpeditor.org)

/*
    Name:
    Author: xiaruize
    Date:
*/
#include <bits/stdc++.h>
using namespace std;
#define int long long
#define ull unsigned long long
#define ALL(a) (a).begin(), (a).end()
#define pb push_back
#define mk make_pair
#define pii pair<int, int>
#define pis pair<int, string>
#define sec second
#define fir first
#define sz(a) int((a).size())
#define rep(i, x, y) for (int i = x; i <= y; i++)
#define repp(i, x, y) for (int i = x; i >= y; i--)
#define Yes cout << "Yes" << endl
#define YES cout << "YES" << endl
#define No cout << "No" << endl
#define NO cout << "NO" << endl
#define debug(x) cerr << #x << ": " << x << endl
#define double long double
const int INF = 0x3f3f3f3f;
const int MOD = 1000000007;
const int N = 2e5 + 10;

// bool st;
int n, a, b;
// bool en;

signed main()
{
    // freopen(".in","r",stdin);
    // freopen(".out","w",stdout);
    // cerr<<(&en-&st)/1024.0/1024.0<<endl;
    // auto t_1=chrono::high_resolution_clock::now();
    cin >> n >> a >> b;
    if (n < a)
        cout << "0" << endl;
    else if (b > a)
        cout << n - a + 1 << endl;
    else
    {
        cout << n / a + (n / a - 1) * (b - 1) + min(n % a, b - 1) << endl;
    }
    // auto t_2=chrono::high_resolution_clock::now();
    // cout <<". Elapsed (ms): " << chrono::duration_cast<chrono::milliseconds>(t_2 - t_1).count() << endl;
    return 0;
}

待更